Puzzles with numbers and things

[snookersfun]

I think I will too! Perhaps a clue, snookersfun?

clue to round 112? It is not a trick question. Definitely doable. Look at all the answers and one of them (by itself) can be easily attributed as a lie. Then take it from there.

Anybody needs to know, which one of the answers, just shout!

[davis_greatest]

I think I will too! Perhaps a clue, snookersfun? 30, 51, 55, 46, 37

[davis_greatest]

lol a tricky one, i'll leave to somelse to answer this

Which one? The one with the children or the counting Fs?

Edit: Oh - just read it properly... it doesn't say they're children

[Sarmu]

lol a tricky one, i'll leave to somelse to answer this

[snookersfun]

and something different totally:

read this:

FINISHED FILES ARE THE RE-
SULT OF YEARS OF SCIENTIF-
IC STUDY COMBINED WITH
THE EXPERIENCE OF YEARS.


Now count the F's out loud in that sentence.
Count them ONLY ONCE!

How many did you all get?

[snookersfun]

round 112 Logic!

...and as I am through one more, I'll pass it on to you. Let's see who'll get to this one. Not much math needed actually, but a good starting point!

Annie, Betty, Carrie, Darla, and Eve recently found out that all of their birthdays were on the same day, though they are different ages. On their mutual birthday, they were jabbering away, flapping their gums about their recent discovery. And, lucky me, I was there. Some of the things that I overheard were...

Darla said to Betty: "I'm nine years older than Eve."
Eve said to Betty: "I'm seven years older than Annie."
Annie said to Betty: "Your age is exactly 70% greater than mine.
Betty said to Carrie: "Eve is younger than you"
Carrie said to Darla: "The difference between our ages is six years."
Carrie said to Annie: "I'm ten years older than you"
Carrie said to Annie: "Betty is younger than Darla."
Betty said to Carrie: "The difference between your age and Darla's is the same as the difference between Darla's and Eve's.

Since I knew these people -- and how old they were, I knew that they were not telling the whole truth. After thinking about it, I realized that when one of them spoke to somebody older than themselves, everything they said was true, but when speaking to somebody younger, everything they said was false.

Now find their ages.

Have fun

Meanwhile round 110 and 111 will be kept open until tomorrow whenever, so if anybody wants to have a go,.... Congrats so far to Abextra for solving round 110 and d_g for solving both of them (figures!)

[snookersfun]

answer to round 108

Barry the Baboon can't wait forever, and nor can I. Here is round 108. Still waiting for answers to be posted here to rounds 106 and 107, so will stop after this round until they appear.

Barry has just made a total clearance (15 reds, each with a colour, followed by the 6 colours).

Fortunately, it was watched by Charlie, who was able to describe the break to me in rather a simpler manner than Barry's descriptions. Charlie tells me that Barry potted every brown left-handed and every blue right-handed. He potted every red left-footed and every pink right-footed. All the other shots, he played with his tail!

Favouring his left side, Barry played more shots left-handed than right-handed, and more shots left-footed than right-footed. However, he scored more points with his tail than he scored left-handed, although not as many as he scored right-handed!


How high was Barry's break?

equation is 5x>7a+3b+2c>4y with y>x

This works out only for x=6 and y=7 (higher than that leaves too many colors considering points needed for the middle equation).
So, one has to fulfill now 7a+3b+2c=29 (8 remaining colours) or maybe easier 7a+3b+2c=17 (4 remaining colours (incl. pink), as one needs a min of 1 colour each), which only works out for 2 additional blacks, 1 add. green and 1 add. pink.

Therefore alltogether 114 points:
1 yellow, 2 greens, 7 browns, 6 blues, 2 pinks, 3 blacks, 15 reds

[snookersfun]

So you mean he paddled at the same speed relative to the flowing water, rather than relative to the river bank?

So, for example, if the river flowed at 1m.ph. and he paddled at 5m.p.h, then relative to the river bank he would travel at 6m.p.h. downstream and 4m.p.h. upstream? (I just made those numbers up - haven't tried the problem yet.)

right, as he paddled at the same speed, he moved at a different speed up- or downstream. So that is a hint!

Anyway meanwhile I had the answer flashing up here, anybody else though?

[davis_greatest]

How fast was the river flowing? (assuming that the vacationer paddled at the same speed all the time)

So you mean he paddled at the same speed relative to the flowing water, rather than relative to the river bank?

So, for example, if the river flowed at 1m.ph. and he paddled at 5m.p.h, then relative to the river bank he would travel at 6m.p.h. downstream and 4m.p.h. upstream? (I just made those numbers up - haven't tried the problem yet.)

River speed is half of the distance travelled before losing his hat divided by the time between losing his hat and turning back... i.e. half of 1 mile in 10 minutes... so 3m.p.h.

[davis_greatest]

Keep them coming, snookersfun - I will resurface soon!

Also, snookersfun or abextra (or anyone else - but I received answers from snookersfun and abextra by Private Message), please post the answer to round 108, while I can still vaguely remember it...

[snookersfun]

I had 2 good answers to round 110 so far, so the next person can answer on the thread.

Meanwhile one more fast one (I know, they are neither new, nor inventive as d_g's, but anything to keep us busy, right?): Round 111

The happy Sunday vacationer rented a paddle boat and had travelled 1 mile up river when his hat blew off (and yes, ignore speed of wind or similar complications). Unconcerned he continued his trip up river, but 10 minutes later he remembered that his return railroad ticket was under the hat band. Turning around immediately he recovered his hat opposite his starting point.

How fast was the river flowing? (assuming that the vacationer paddled at the same speed all the time)

[snookersfun]

round 110

one more on special request

A young boy, his father and his mother were celebrating a birthday.
This got the young boy to thinking about their ages. Upon asking
his father about this, the father told him: "Well, right now I am
six times your age, and all together the sum of our ages (yours,
mine, and your mother's) is 70. Later on, when I am only twice
your age, that sum will be twice what it is now."

Who's birthday is it?

[snookersfun]

of course! shouldn't try to post and give homework help at the same time...

[davis_greatest]

Now the times for each person to travel the respective parts until noon are
x/4y and y/9x respectively, which of course have to be equal.

You have these the wrong way up - the times should be 4y/x and 9x/y, but yes, they are equal, and the rest is OK. Nearly as elegant as my way.

[snookersfun]

... and that of course is the right answer! More fame and all for you d_g!

I had done it slightly different and just put that way up as well (as very elegant):

I divided the distance traveled into x and y (before/after noon).
Therefore speed of one person x/4 and second y/9 unit distances/h.
Now the times for each person to travel the respective parts until noon are
x/4y and y/9x respectively, which of course have to be equal.

x/4y=y/9x
or x²=4y²/9, or x=⅔ y

so distance x is ⅔ of y and as the faster person needs 4 hours to travel distance y, he needs 6 hours for distance x...